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    I came across a maths question in an OCR C4 paper that at one point involved integrating Cos22x.

    The two identities regarding Cos2x i know of are:
    cos2x = 2cos2x - 1 and cos2x= 1-2sin2x

    I tried squaring both of those identites but ultimately got nothing that seems like it could be easily integrated eg. 4cos4x-4cos2x+1

    The mark scheme says Cos22x = 1/2 (1+cos4x) (which can be easily integrated)

    I was just curious to know how they got to 1/2 (1+cos4x), in case a similar question was to come up in the exam.
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    (Original post by MohammedPatel)
    I came across a maths question in an OCR C4 paper that at one point involved integrating Cos22x.

    The two identities regarding Cos2x i know of are:
    cos2x = 2cos2x - 1 and cos2x= 1-2sin2x

    I tried squaring both of those identites but ultimately got nothing that seems like it could be easily integrated eg. 4cos4x-4cos2x+1

    The mark scheme says Cos22x = 1/2 (1+cos4x) (which can be easily integrated)

    I was just curious to know how they got to 1/2 (1+cos4x), in case a similar question was to come up in the exam.
    Let A = 2x. This way Cos2A = 2cos^2(A) - 1. Now sub back A = 2x. Can you see it now?
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    (Original post by MohammedPatel)
    I came across a maths question in an OCR C4 paper that at one point involved integrating Cos22x.

    The two identities regarding Cos2x i know of are:
    cos2x = 2cos2x - 1 and cos2x= 1-2sin2x

    I tried squaring both of those identites but ultimately got nothing that seems like it could be easily integrated eg. 4cos4x-4cos2x+1

    The mark scheme says Cos22x = 1/2 (1+cos4x) (which can be easily integrated)

    I was just curious to know how they got to 1/2 (1+cos4x), in case a similar question was to come up in the exam.
    \cos 2\theta = 2\cos^2 \theta - 1

    Let \theta = 2x

    Then \cos (4x) = 2\cos^2 (2x) - 1 \Rightarrow \cos^2 2x = \cdots
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    Thanks guys
 
 
 
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